Lesson Explainer: Atmospheric Pressure Physics

In this explainer, we will learn how to describe atmospheric pressure using various units, including the height of a mercury column.

When looking at a column of fluid, whether a gas or a liquid, the pressure can be calculated as follows.

Equation: Pressure in a Column of Fluid

The equation used to describe the pressure exerted by the weight of a gas or liquid in a column is 𝑃=πœŒπ‘”β„Ž, where 𝑃 is the pressure, 𝜌 is the density of the fluid, 𝑔 is the acceleration due to gravity (9.81 m/s2 for Earth), and β„Ž is the height above the point we are looking at.

A column of water is usually a tube full of water, and the pressure it exerts on a point can be seen to increase when traveling deeper, since height β„Ž is larger. This same pressure also exists for the columns of atmosphere, with the top of the column being placed at the very top of the atmosphere of Earth.

Looking at the diagram above, when comparing the pressure of the same heights of water and atmosphere in a column (comparing yellow points at the bottom), the water exerts a higher pressure because it has a higher density.

The atmosphere still applies a pressure, which was first measured by the Italian scientist Torricelli. This was done by placing an inverted tube of mercury into a pool of mercury, then letting the tube drain into the pool, as shown below.

All the way from the top of the atmosphere down to the apparatus exerts a pressure onto the surface of the mercury, which pushes it up the tube, while the mercury in the tube is pulled down by gravity. The forces balance out to make the mercury in the tube fall, but not all the way.

For a 1-metre-long tube at sea level, the distance between the top of the pool of mercury and the top of the mercury in the tube is 760 mm.

The atmospheric pressure changes based on elevation. Being higher up means less pressure, as there is less air.

Let’s look at some examples.

Example 1: Region Left after Mercury Drop

The apparatus shown in the diagram is used to measure atmospheric pressure. Which of the following occupies region A of the test tube?

  1. Mercury vapor
  2. Vacuum
  3. Glass
  4. Air


When the mercury tube is placed into the pool of mercury, it is completely full of mercury. It is still full of mercury when the bottom is opened, releasing it into the pool, meaning there is nothing to replace the space created in region A.

While there may be small amounts of residual mercury, region A is predominantly a vacuum. It was a full space, and now it is not. Torricelli’s experiment was the earliest artificial vacuum.

The correct answer is B, vacuum.

Example 2: Atmospheric Pressure at Different Elevations

The apparatus shown in the diagram is used to measure atmospheric pressure. In which case is the apparatus at the greatest height above sea level?

  1. I
  2. II
  3. III
  4. There is no difference in the apparatus’s height above sea level in the three cases.


A greater atmospheric pressure means a higher force pressing down on the mercury pool surface. This would result in pushing the mercury in the tube up higher, so apparatus I has the highest atmospheric pressure, and III has the lowest.

Let’s look at the equation for pressure: 𝑃=πœŒπ‘”β„Ž.

Pressure is related to the density of the fluid, gravity, and height above a particular point. An increase in elevation would correspond to a decrease in the height as you get closer to the edge of the atmosphere. So higher elevations mean lower atmospheric pressures.

The apparatus with the greatest height above sea level would thus be the one with the least atmospheric pressure pressing down on it, which is apparatus III. The answer is thus C.

Atmospheric pressure has several ways it can be expressed in units, as seen in the table below.

Standard AtmospherePascalBarTorrMillimeters of Mercury
1 atm101β€Žβ€‰β€Ž325 Pa1.01 bar760 torr760 mmHg

Pascal (Pa) is the SI unit, with 1 pascal being equal to 1 newton per square metre:1=1.PaNm

The atmosphere of Earth, called the standard atmosphere (atm), is also a unit. The atmospheric pressure of Earth is 1 atm. One atm is 101β€Žβ€‰β€Ž325 Pa, often shortened to 101 kPa: 1=101325.atmPa

Another unit is used for even further shortening, the bar. 1 bar (also abbreviated as β€œbar”) is equal to 100β€Žβ€‰β€Ž000 Pa, so it is often used to shorten pascals. One atm is thus 1.01 bar:1=100000.barPa

The last two units of pressure, torr and millimetres of mercury, have a 1 to 1 ratio. One torr is one millimetre of mercury, and 760 torr is one atm:1=1,1=760.torrmmHgatmtorr

Let’s look at some examples.

Example 3: Bar to Pascal Conversion

A bar is a unit that is defined as being equal to 10 Pa. Convert a pressure of 0.48 bar to a pressure in pascals.

  1. 4.8Γ—10
  2. 4.8Γ—10οŠͺ
  3. 4.8Γ—10
  4. 4.8Γ—10
  5. 4.8Γ—10οŠͺ


We know that 1=100000.barPa

Or, in other words, there are 100β€Žβ€‰β€Ž000 pascals per 1 bar:1000001.Pabar

So if there is only 0.48 bar, we just need to multiply it with this relation:1000001Γ—0.48.Pabarbar

The units of bar cancel, leaving just Pa. We thus have100000Γ—0.48=48000.PaPa

So, 0.48 bar is 48β€Žβ€‰β€Ž000 pascals, equivalent to 4.8Γ—10οŠͺ.

The correct answer is thus E.

Example 4: Torr Conversions

A torr is a unit that is defined as being equal to the pressure produced by 1 mmHg.

  1. Convert a pressure of 455 torr to a pressure in pascals.
    1. 6.06Γ—10
    2. 4.55Γ—10
    3. 6.37Γ—10
    4. 6.19Γ—10οŠͺ
    5. 6.06Γ—10οŠͺ
  2. Convert a pressure of 455 torr to a pressure in bar.
    1. 0.606 bar
    2. 4.55 bar
    3. 6.37 bar
    4. 0.637 bar
    5. 0.619 bar


Part 1

To convert torr to pascal, let’s relate them both to the standard atmosphere, atm:1=760,1=101325.atmtorratmPa

We can then relate the conversions as follows: 760=101325.torrPa

Or, in other words, for every 101β€Žβ€‰β€Ž325 pascals, there are 760 torr:101325760.Patorr

If we multiply this relation by 455 torr, we can cancel the units: 101325760Γ—455.Patorrtorr

Working out the numbers gives 101325760Γ—455=60661.PaPa

In scientific notation, 60β€Žβ€‰β€Ž661 Pa becomes 6.06Γ—10οŠͺ. The answer is thus E.

Part 2

The subsequent conversion to bar is easy now that we have it in pascals. 1 bar is 100β€Žβ€‰β€Ž000 Pa:1=100000.barPa

So, for every 1 bar, there is 100β€Žβ€‰β€Ž000 pascals:1100000.barPa

Let’s multiply this relation by the pressure in pascals, 60β€Žβ€‰β€Ž661 Pa:1100000Γ—60661.barPaPa

Units of pascal cancel, giving 1100000Γ—60661=0.606,barbar which means 455 torr in bar is 0.606 bar. The answer is thus A.

Now that we have seen how to convert between units of atmospheric pressure, let’s look back at the original equation for calculating pressure:𝑃=πœŒπ‘”β„Ž.

There are many different units of pressure, and those units can have prefixes that change them further, such as using kilopascal (kPa), or measuring using centimetre of mercury (cmHg), instead of millimetres of mercury (mmHg). Most times, the final result will be in pascals when using this equation, but always be wary of the units used.

Let’s look at an example.

Example 5: Finding the Upward Pressure on a Mercury Column

The apparatus shown in the diagram is used to measure atmospheric pressure. Find the upward pressure on the mercury column. Use a value of 13β€Žβ€‰β€Ž595 kg/m3 for the density of mercury.

  1. 133 kPa
  2. 175 kPa
  3. 203 kPa
  4. 50.6 kPa
  5. 101 kPa


The upward pressure inside the column of mercury depends on its height above the pool, which is thankfully given, 0.76 m. In the equation for pressure, this is the value of β„Ž:𝑃=πœŒπ‘”β„Ž.

The force of gravity on Earth, 𝑔, is 9.81 m/s2. We are also given the density 𝜌, which is 13β€Žβ€‰β€Ž595 kg/m3. We thus have all the variables we need to fit into the equation for pressure:𝑃=ο€Ή13595/9.81/(0.76).kgmmsm

The units of metres in gravity and height cancel with the ones given by density. Multiplying everything through gives 𝑃=101358Γ—.kgms

Pascals are equivalent to newtons per square metre (N/m2) and newtons are equivalent to kilogram-metres per second squaredο€ΉΓ—/kgms. We can convert the units we have to newtons per square metre as follows:kgmskgmsmkgmsmNmΓ—=×1οˆο€½Γ—ο‰ο€Ό1=.

Thus, the final pressure is 101358Γ—=101358.kgmsPa

101β€Žβ€‰β€Ž358 Pa is 101 kPa. The correct answer is thus E.

Let’s summarize what we have learned in this explainer.

Key Points

  • The equation to describe the pressure of a liquid or gas is 𝑃=πœŒπ‘”β„Ž, where 𝑃 is the pressure, 𝜌 is the density of the fluid, 𝑔 is the force due to gravity, and β„Ž is the height.
  • A column of mercury can be used to measure atmospheric pressure.
  • 1 atmΒ =Β 101β€Žβ€‰β€Ž325 PaΒ =Β 1.01 bar
  • 1 atmΒ =Β 760 mmHgΒ =Β 760 torr

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